Please help me prove nlogn -n belongs to Omega nlogn - algorithm

I try to solve this problem many times but I still don't have a solution.
The question as follows:
Use the definition of Big Omega to prove that
nlogn - n belongs to Omega of nlogn.
Thank you

As follows:
f(n) = n log n - n
>= n log n - 0.5 n log n, for all n > 100 (assuming log10 base)
= 0.5 * n log n,
= c * n log n, for c = 0.5, n > n0 = 100
To summarize:
f(n) >= c * g(n), for all n > n0 = 100, and
with c = 0.5, and
with g(n) = n log n.

When logn > 2, then 1/2 nlogn - n > 0. Adding 1/2 nlogn to both sides gives nlogn - n > 1/2 nlogn.
This proves that nlogn - n is Omega(nlogn) (with the constant 1/2, and N the smallest integer such that logN>2).

Related

Asymptotic notation Big Omega analysis

I've tried calculating this function and I am a bit unsure of my result. I set it to True. Can anyone explain if my answer is correct and why?
(3 log 2 n + 55 log(n 10 ) + 8 log n) · log n = Ω(log 10 n)
I set it to True
Your result is correct, but can be further simplified to Ω(log(n)) as log(10n) + log(10) + log(n) and log(10) is a constant.
To prove that f(n) = Ω(g(n)) you need to show that g(n) is a "lower bound" asymptotically of f(n).
The formal definition is that f(n) = Ω(g(n)) is there exists some c, n0 > 0 s.t. for all n > n0 it holds that f(n) >= g(n).
Recall that for every natural integer bigger than 2 it holds that log(n) > 1 so
(3log(2n) + 55log(10n) + 8log n) · log n > 3log(2n) + 55log(10n) + 8log(n) > 8log(n) > log(n).
Choose c = 1, n0 = 2 and we got that for all n > n0: (3log(2n) + 55log(10n) + 8log n) · log n > log(n), thus (3log(2n) + 55log(10n) + 8log n) · log n = Ω(log(n).

which algorithm dominates f(n) or (g(n)

I am doing an introductory course on algorithms. I've come across this problem which I'm unsure about.
I would like to know which of the 2 are dominant
f(n): 100n + log n or g(n): n + (log n)^2
Given the definitions of each of:
Ω, Θ, O
I assumed f(n), so fn = Ω(g(n))
Reason being that n dominates (log n)^2, is that true?
In this case,
limn → ∞[f(n) / g(n)] = 100.
If you go over calculus definitions, this means that, for any ε > 0, there exists some m for which
100 (1 - ε) g(n) ≤ f(n) ≤ 100 (1 + ε) g(n)
for any n > m.
From the definition of Θ, you can infer that these two functions are Θ of each other.
In general, if
limn → ∞[f(n) / g(n)] = c exists, and
0 < c < ∞,
then the two functions have the same order of growth (they are Θ of each other).
n dominates both log(n) and (log n)^2
A little explanation
f(n) = 100n + log n
Here n dominates log n for large values of n.
So f(n) = O(n) .......... [1]
g(n) = n + (log n)^2
Now, (log n)^2 dominates log n.
But n still dominates (log n)^2.
So g(n) = O(n) .......... [2]
Now, taking results [1] and [2] into consideration.
f(n) = Θ(g(n)) and g(n) = Θ(f(n))
since they will grow at the same rate for large values of n.
We can say that f(n) = O(g(n) if there are constants c > 0 and n0 > 0 such that
f(n) <= c*g(n), n > n0
This is the case for both directions:
# c == 100
100n + log n <= 100(n + (log n)^2)
= 100n + 100(log(n)^2) (n > 1)
and
# c == 1
n + (log n)^2 <= 100n + log n (n > 1)
Taken together, we've proved that n + (log n)^2 <= 100n + log n <= 100(n + (log n)^2), which proves that f(n) = Θ(g(n)), which is to say that neither dominates the other. Both functions are Θ(n).
g(n) dominates f(n), or equivalently, g(n) is Ω(f(n)) and the same hold vice versa.
Considering the definition, you see that you can drop the factor 100 in the definition of f(n) (since you can multiply it by any fixed number) and you can drop both addends since they are dominated by the linear n.
The above follows from n is Ω(n + logn) and n is Ω(n + log^2n.
hope that helps,
fricke

Algorithm's complexity : Big O notation

O( sqrt(n) ) = O(n) ?
We should find c and n0 verifying :
0 ( sqrt(n) ) < c*n ; c>0 and n>n0
How to find c and n0 ? or should i find another idea ?
Thanks
For n > 1, we have √n > 1, hence we have the following inequality:
√n < √n * √n = n, for any n > 1.
So we can take c = 1 and n0 = 2 to prove that √n = O(n).
Remark
Strictly speaking, you should avoid writing down something like O(√n) = O(n). Big-O notation is for describing the asymptotic upper bound of a function, but O(√n) is not a function.
O(√n) = O(n) is an abuse of notation, it really means the following:
If f is a function such that f(n) = O(√n), then f(n) = O(n).
In our case, if for any function f we have f(n) = O(√n), since √n < n for any n > 1, clearly we have f(n) < c * √n < c * n for any n > 1, and hence f(n) = O(n).

Solving Recurrence relation: T(n) = 3T(n/5) + lgn * lgn

Consider the following recurrence
T(n) = 3T(n/5) + lgn * lgn
What is the value of T(n)?
(A) Theta(n ^ log_5{3})
(B) Theta(n ^ log_3{5})
(c) Theta(n Log n )
(D) Theta( Log n )
Answer is (A)
My Approach :
lgn * lgn = theta(n) since c2lgn < 2*lglgn < c1*lgn for some n>n0
Above inequality is shown in this picture for c2 = 0.1 and c1 = 1
log_5{3} < 1,
Hence by master theorem answer has to be theta(n) and none of the answers match. How to solve this problem??
Your claim that lg n * lg n = Θ(n) is false. Notice that the limit of (lg n)2 / n tends toward 0 as n goes to infinity. You can see this using l'Hopital's rule:
limn → ∞ (lg n)2 / n
= lim n → ∞ 2 lg n / n
= lim n → ∞ 2 / n
= 0
More generally, using similar reasoning, you can prove that lg n = o(nε) for any ε > 0.
Let's try to solve this recurrence using the master theorem. We see that there are three subproblems of size n / 5 each, so we should look at the value of log5 3. Since (lg n)2 = o(nlog5 3), we see that the recursion is bottom-heavy and can conclude that the recurrence solves to O(nlog5 3), which is answer (A) in your list up above.
Hope this helps!
To apply Master Theorem we should check the relation between
nlog5(3) ~= n0.682 and (lg(n))2
Unfortunately lg(n)2 != 2*lg(n): it is lg(n2) that's equal to 2*lg(n)
Also, there is a big difference, in Master Theorem, if f(n) is O(nlogb(a)-ε), or instead Θ(nlogba): if the former holds we can apply case 1, if the latter holds case 2 of the theorem.
With just a glance, it looks highly unlikely (lg(n))2 = Ω(n0.682), so let's try to prove that (lg(n))2 = O(n0.682), i.e.:
∃ n0, c ∈ N+, such that for n>n0, (lg(n))2 < c * n0.682
Let's take the square root of both sides (assuming n > 1, the inequality holds)
lg(n) < c1 * n0.341 , (where c1 = sqrt(c))
now we can assume, that lg(n) = log2(n) (otherwise the multiplicative factor could be absorbed by our constant - as you know constant factors don't matter in asymptotic analysis) and exponentiate both sides:
2lg(n) < 2c2 * n0.341 <=> n < 2c2 * n0.341 <=> n < (n20.341)c2 <=> n < (n20.341)c2 <=> n < (n1.266)c2
which is immediately true choosing c2 = 1 and n0 = 1
Therefore, it does hold true that f(n) = O(nlogb(a)-ε), and we can apply case 1 of the Master Theorem, and conclude that:
T(n) = O(nlog53)
Same result, a bit more formally.

Is O(log(n*log n) can be considered as O(log n)

Consider I get f(n)=log(n*log n). Should I say that its O(log(n*log n)?
Or should I do log(n*log n)=log n + log(log n) and then say that the function f(n) is O(log n)?
First of all, as you have observed:
log(n*log n) = log(n) + log(log(n))
but think about log(log N) as N->large (as Floris suggests).
For example, let N = 1000, then log N = 3 (i.e. a small number) and log(3) is even smaller,
this holds as N gets huge, i.e. way more than the number of instructions your code could ever generate.
Thus, O(log(n * log n)) = O(log n + k) = O(log(n)) + k = O(log n)
Another way to look at this is that: n * log n << n^2, so in the worse case:
O(log(n^2)) > O(log(n * log n))
So, 2*O(log(n)) is an upper bound, and O(log(n * log n)) = O(log n)
Use the definition. If f(n) = O(log(n*log(n))), then there must exist a positive constant M and real n0 such that:
|f(n)| ≤ M |log(n*log(n))|
for all n > n0.
Now let's assume (without loss of generality) that n0 > 0. Then
log(n) ≥ log(log(n))
for all n > n0.
From this, we have:
log(n(log(n)) = log(n) + log(log(n)) ≤ 2 * log(n)
Substituting, we find that
|f(n)| ≤ 2*M|log(n))| for all n > n0
Since 2*M is also a positive constant, it immediately follows that f(n) = O(log(n)).
Of course in this case simple transformations show both functions differ by a constant factor asymptotically, as shown.
However, I feel like it is worthwhile remind a classic test for analyzing how two functions relate to each other asymptotically. So here's a little more formal proof.
You can check how does f(x) relates to g(x) by analyzing lim f(x)/g(x) when x->infinity.
There are 3 cases:
lim = infinty <=> O(f(x)) > O(g(x))
inf > lim > 0 <=> O(f(x)) = O(g(x))
lim = 0 <=> O(f(x)) < O(g(x))
So
lim ( log( n * log(n) ) / log n ) =
lim ( log n + log log (n) ) / log n =
lim 1 + log log (n) / log n =
1 + 0 = 1
Note: I assumed log log n / log n to be trivial but you can do it by de l'Hospital Rule.

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